Q78 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
Suppose $A$ and $B$ are events of a random experiment such that $P(A)=\frac{1}{3}, P(A \cap B)=\frac{1}{5}$ and $P(A \cup B)=\frac{3}{5}$.
Then, match the items of List-I with the items of List-II
- A(a)
- B(b)
- C(c)
- Correct answerD(d)
Solution
Given that,
$$
P(A)=\frac{1}{3}, P(A \cap B)=\frac{1}{5}, P(A \cup B)=\frac{3}{5}
$$
We know that,
$$
\begin{gathered}
P(A \cup B)=P(A)+P(B)-P(A \cup B) \\
\frac{3}{5}=\frac{1}{3}+P(B)-\frac{1}{5} \\
P(B)=\frac{3}{5}+\frac{1}{5}-\frac{1}{3}=\frac{9+3-5}{15} \Rightarrow P(B)=\frac{7}{15}
\end{gathered}
$$
(a) $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{5}}{\frac{7}{15}}=\frac{3}{7}$
(b) $\quad P(\bar{B})=1-P(B)=1-\frac{7}{15}=\frac{8}{15}$
(c) $P(A \cap \bar{B})=P(A)-P(A \cap B)=\frac{1}{3}-\frac{1}{5}=\frac{5-3}{15}=\frac{2}{15}$
(d) $P(B \cap \bar{A})=P(B)-P(A \cap B)=\frac{7}{15}-\frac{1}{5}=\frac{7-3}{15}=\frac{4}{15}$
Hence, the correct answer is option: 'D'.